Problem: $ \left(\dfrac{81}{16}\right)^{-\frac{3}{4}}$
Answer: $= \left(\dfrac{16}{81}\right)^{\frac{3}{4}}$ $= \left(\left(\dfrac{16}{81}\right)^{\frac{1}{4}}\right)^{3}$ To simplify $\left(\dfrac{16}{81}\right)^{\frac{1}{4}}$ , figure out what goes in the blank: $\left(? \right)^{4}=\dfrac{16}{81}$ To simplify $\left(\dfrac{16}{81}\right)^{\frac{1}{4}}$ , figure out what goes in the blank: $\left({\dfrac{2}{3}}\right)^{4}=\dfrac{16}{81}$ so $ \left(\dfrac{16}{81}\right)^{\frac{1}{4}}=\dfrac{2}{3}$ So $\left(\dfrac{16}{81}\right)^{\frac{3}{4}}=\left(\left(\dfrac{16}{81}\right)^{\frac{1}{4}}\right)^{3}=\left(\dfrac{2}{3}\right)^{3}$ $= \left(\dfrac{2}{3}\right)\cdot\left(\dfrac{2}{3}\right)\cdot \left(\dfrac{2}{3}\right)$ $= \dfrac{4}{9}\cdot\left(\dfrac{2}{3}\right)$ $= \dfrac{8}{27}$